perl - File::Basename-fileparse returns "File::Basename"

Question

For some reason my code is doing this wierd thing where `fileparse` only prints (literally) `File::Basename` ``` use strict; use warnings 'all'; use File::Basename; ... my $fileName = File::Basename->fileparse($filePath); print("$filePath
"); print("$fileName
"); ``` And output is: ``` a/b/c/d.bin File::Basename ``` What did I do wrong?

Accepted Answer

The `fileparse` is not a method; it is a function. This function is exported by default, so you actually want to do

```
use File::Basename;
my $fileName = fileparse($filePath);
```

You have used is as a method call. Here `File::Basename->fileparse($filePath)` is equivalent to

```
fileparse("File::Basename", $filePath)
```

because in a method invocation, the invocant (usually an object; here the package name) becomes the first argument. This is wrong, as it treats `"File::Basename"` as the path to parse, and the following arguments as a list of valid suffixes.

If you want to use the `fileparse` function without exporting it to your namespace, you could

```
use File::Basename (); # note empty parens that supress the import
File::Basename::fileparse(...); # use fully qualified name
```

perl - File::Basename->fileparse returns "File::Basename"

Solution

Problem